Answer
$$\frac{1}{3}{\left( {\sec x} \right)^3} + C $$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^3}x} \tan xdx \cr
& {\text{We write }}{\sec ^3}x{\text{ as }}{\sec ^2}x\sec x \cr
& = \int {{{\sec }^2}x\sec x} \tan xdx \cr
& {\text{Integrate by substitution method}} \cr
& {\text{Let }}u = \sec x,\,\,\,\,du = \sec x\tan xdx,\,\,\,\,dx = \frac{{du}}{{\sec x\tan x}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {{{\sec }^2}x\sec x} \tan xdx = \int {{u^2}\sec x\tan x\left( {\frac{{du}}{{\sec x\tan x}}} \right)} \cr
& {\text{Cancel common factor }}\sec x\tan x \cr
& = \int {{u^2}} du \cr
& {\text{integrating}} \cr
& = \frac{1}{3}{u^3} + C \cr
& {\text{write in terms of }}x,\,\,\,u = \sec x \cr
& = \frac{1}{3}{\left( {\sec x} \right)^3} + C \cr} $$
